Question

The hybridisation in $I{F}_7$ molecule is :
A. $s{p}^3$
B. $s{p}^3{d}^2$
C. $s{p}^{3}d$
D. $s{p}^3{d}^3$

Answer 1

Hint: In the process of hybridization two or more than two atomic orbital goes in inter-mixing of the atomic orbital and gives hybridized orbital which remains same in the number and energy.

Complete step by step solution:
> The given compound is known as iodine heptafluoride. It has seven bond pairs and one lone pair. In the following diagram pentagonal bipyramidal geometry of molecules is shown . > There are two fluorine atoms above and below the plane which make two pyramids with the plane while five fluorine atoms are arranged in a pentagon in the plane. The geometry of molecules used to predict with VSEPR theory (valence shell electron repulsion theory).In this theory the number of electron pairs are used to predict the geometry of molecules.



> In $I{F}_7$ molecule ,central metal atom is $I$. The electronic configuration of iodine atom is given below ;
$I_{53}=1s^2,2s^2,2p^6,3s^2,3p^6,4s^2,3d^{10},4p^6,5s^2,4d^{10},5p^5$ so here is seven atomic orbital that is one s orbitals, three p orbitals and three d orbitals which hybridizes to form seven hybrid orbitals. The geometry of $I{F}_7$ is pentagonal bipyramidal. Seven hybrid $s{p}^3{d}^3$ orbitals are directed towards the corners of a pentagonal bipyramid. Five of the hybrid orbitals are directed towards the corners of a regular pentagon whereas the remaining two are directed above and below the plane. So here option D is correct.

Note : We know that the number of valence electron in central metal atom Iodine is seven and in fluorine it is also seven so total number of valence electron in molecule is $7+7\times 7=56$.Hence it need seven atomic orbital so possess $s{p}^3{d}^3$ hybridization. The bond angle is ${72}^0$ and ${90}^0$.